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I should find for which x:es this function is defined:

$$ f(x) = \arcsin\frac{x}{\sqrt{1+x^2}} $$

How would one go about that?

I can "see" that this is defined on the interval:

$$ (-\infty, \infty) $$

so

$$ D_f = (-\infty, \infty) $$

because the quotient is always between $$ [-1, 1] $$

So, how would I mathematically prove this?

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1 Answer 1

up vote 3 down vote accepted

Well you can define $f$ for all $x\in \mathbb R$, where $\sqrt{1+x^2} \neq 0$ (because dviding by $0$ is not defined, and all other operations are allowed for all $x \in \mathbb R$).

Following that up, we get that $f$ is defined at all $x \in \mathbb R$, where $x^2 \neq -1$. So for this part, all $x \in \mathbb R$ are allowed.

As Chris pointed out, we have to check for $\arcsin (z)$, where $z = \frac{x}{\sqrt{x^2+1}}$. We know that $\arcsin(z)$ is only defined for $z \in [-1,1]$. Let's show $|z|\leq 1$, if $z = \frac{x}{\sqrt{x^2+1}}$. As $|x| = \sqrt{x^2} <\sqrt{x^2+1}$, and $\sqrt{x^2+1} > 0 \forall x\in \mathbb R$, we have $|z| = |\frac{x}{\sqrt{x^2+1}}|<1 \forall x \in \mathbb R$.

Because of this, $D_f = \mathbb R$.

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$\arcsin(x)$ is not defined for all $x$. –  Chris Eagle Oct 11 '12 at 16:00
    
Oh I total didn't see arcsin. Will edit. –  Stefan Oct 11 '12 at 16:00
    
Thank you, Stefan! :-) –  Curtain Oct 11 '12 at 16:15
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